Solve this equation $\log_2x=\log_{5-x}3$ Solve this equation
$$\displaystyle \log_2x=\log_{5-x}3$$
the answer is $x=2,x=3$
http://www.wolframalpha.com/input/?i=log_2%28x%29%3Dlog%285-x%2C3%29
Can you give me some hint
 A: It’s certainly at least worth trying to see what value of $x$ makes the arguments of the logs equal and what value makes the bases equal, and when it turns out that the same value, $x=3$, does both, you’re done: the two expressions are then identical.
Added: Having spotted one trivial solution, it’s worth thinking about whether there are others. Are there any values of $x$ that make either side especially simple? If $x=2$, the lefthand side is $\log_22$, which we can immediately evaluate exactly as $1$. And pleasantly enough, it just happens that the righthand side is then $\log_33$, which of course is also $1$.
A generally solution is hard. If you have the equation $\log_b x=\log_{a-x}c$, you can reduce both sides to the same base. If you pick natural logs, you use the fact that $$\log_bx=\frac{\ln x}{\ln b}$$ to write $$\frac{\ln x}{\ln b}=\frac{\ln c}{\ln(a-x)}\;,$$ or $\ln x\ln(a-x)=\ln b\ln c$. This is very ugly in general, but note that $x+(a-x)=a$; if you’re lucky enough to find that $b+c=a$, as in this problem, you clearly get two solutions by setting $x=b$ and $x=c$.
A: $\log_2x=\log_{5-x}3$ 
$$\implies \frac{\log x}{\log2} = \frac{\log3}{\log (5-x)} $$
$$ \implies \log x \cdot \log(5-x) = \log2 \cdot \log3$$
$$\implies \log\log(5-x)^x = \log\log(3)^2$$
or $$\log\log(5-x)^x =\log\log(2)^3$$
Comparing both, we get either $x = 2$ , or $x = 3$. 
A: By the above equation,
Log2*log3=logx*log(5-x)
Now x must be between 5 and 0. Seeing some logical cases, 5-x is 3 or x is 3. Hence you get the answers above. Do i have to prove there is no other complex x for which this equation holds true?
I do not think there will be a complex answer, since x can only lie between 0 and 5. One thing we should try is log2/log(5-x)=logx/log3, but i do not thing it will yield anything. Since log2*log3 is  constant, the set of possible x should be very limited.
